A part of a complete circuit is shown in the figure. At some instant, the value of current I is 1A and it is decreasing at a rate
of 10² A s –1 . The value of the potential difference V P – V Q , (in volts) at that instant, is _________.

Solution

Physics sir · 4 steps

Walk from Q to P along the current and add up each drop; let the inductor's sign follow the falling current.

1. Read the figure. The arrow shows I flowing from Q towards P. It enters the battery at the short thick plate (negative) and leaves at the long thin plate (positive). So walking from Q to P you follow the current the whole way.
2. Each element. Across R the potential falls by IR = 1 × 2 = 2 V. Through the battery, from - to +, it rises by 30 V. Across the inductor it changes by -LdI/dt, and here dI/dt = -10² A s⁻¹ because the current is falling.

   V_Q - IR + 30 - LdI/dt = V_P

3. Inductor term. -LdI/dt = -(50 × 10⁻³) × (-100) = +5 V. The falling current makes the inductor act like a small cell pushing current forward, so P sits 5 V higher than the inductor's other end.
4. So. V_P - V_Q = -2 + 30 + 5 = 33 V.

The common slip: 23. Taking dI/dt as +100 instead of -100 makes the inductor a 5 V drop and gives 30 - 2 - 5 = 23, not 33.

The idea to keep. Treat an inductor like a resistor-style drop of LdI/dt along the current. If the current is decreasing, that drop turns into a rise, because the inductor tries to keep the current going.

Answer from the NTA final answer key. Solution by PrepWiser.